MG ENGINEERING ECONOMICS AND COST ANALYSIS. L T P C. 3 0 0 3 . The issues that are covered in this book are elementary economic analysis. Engineering Economics. Cash Flow FERC. a1. Engineering Economics . What is the book value of the asset in the previous example after 3 years. Engineering economic analysis / Donald G. Newnan, Ted G. Eschenbach, Jerome . This book is designed to teach the fundamental concepts of engineering.
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Download Engineering Economics By R. Panneerselvam – Designed as a text book for undergraduate students in various engineering disciplines – mechanical . This book on Engineering Economics is the outgrowth of my several years of teaching postgraduate courses in industrial engineering and production. Engineering Economics. Principles. ▫ During our examinations we assume a consolidated economy. (Free of extremities, such as war, hyperinflation, corruption.
What are the ways by which the economic efficiency can be improved? August It is often of an emergency nature which will have associated penalty in terms of expediting cost of maintenance and down time cost of equipment. Therefore, machine A should be selected. Then, in the second step, the annual equivalent cost is computed using the following equation: The process sequence of a component which has been planned in the past is not static. Beyond that point, the cost of preventive maintenance will be more when compared to the breakdown maintenance cost.
Interest Formulas and Their Applications 4. Present Worth Method of Comparison 5. Future Worth Method 6. Annual Equivalent Method 7. Rate of Return Method 8. Replacement and Maintenance Analysis 9. Depreciation Evaluation of Public Alternatives Inflation Adjusted Decisions Nondeterministic Decision Making Cash Flows with Probability Distributions Selection of Alternatives Using Simulation Inventory Control Make or Buy Decision If the firm must select one of the materials handling systems, which one is the most desirable?.
A firm has identified three mutually exclusive alternatives. The life of all three alternatives is estimated to be five years. Find the best alternative based on the rate of return method. An automobile company is planning to buy a robot for its forging unit. It has identified two different companies for the supply of the robot. The details of cost and incremental revenue of using robots are summarized in the following table: Brand Speedex Giant Initial cost Rs.
Suggest the best brand of robot to the company based on the rate of return method. A bank introduces two different investment schemes whose details are as follows: A company is planning for its expansion programme which will take place after five years. The expansion requires an equal sum of Rs. Gamma Bank has recently introduced a scheme in this line.
If the company invests Rs. Suggest whether the company should invest with the Gamma Bank for its expansion programme. Consider the following table which summarizes data of two alternatives.
First cost Annual return Life Alternative 1 Rs. A company is planning to expand its present business activity. It has two alternatives for the expansion programme and the corresponding cash flows are given in the following table. In addition to these facilities, there are several other items which are necessary to facilitate the functioning of organizations.
All such facilities should be continuously monitored for their efficient functioning; otherwise, the quality of service will be poor. Besides the quality of service of the facilities, the cost of their operation and maintenance would increase with the passage of time. Hence, it is an absolute necessity to maintain the equipment in good operating conditions with economical cost. Thus, we need an integrated approach to minimize the cost of maintenance.
In certain cases, the equipment will be obsolete over a period of time. If a firm wants to be in the same business competitively, it has to take decision on whether to replace the old equipment or to retain it by taking the cost of maintenance and operation into account. There are two basic reasons for considering the replacement of an equipment—physical impairment of the various parts or obsolescence of the equipment.
Physical impairment refers only to changes in the physical condition of the machine itself. This would lead to a decline in the value of the service rendered, increased operating cost, increased maintenance cost or a combination of these. Obsolescence is due to improvement of the tools of production, mainly improvement in technology.
So, it would be uneconomical to continue production with the same machine under any of the above situations. Hence, the machines are to be periodically replaced. Sometimes, the capacity of existing facilities may be inadequate to meet the current demand.
Under such situation, the following alternatives will be considered. Preventive maintenance PM is the periodical inspection and service activities which are aimed to detect potential failures and perform minor adjustments or repairs which will prevent major operating problems in future. Breakdown maintenance is the repair which is generally done after the equipment has attained down state. It is often of an emergency nature which will have associated penalty in terms of expediting cost of maintenance and down time cost of equipment.
Preventive maintenance will reduce such cost up to a point. Beyond that point, the cost of preventive maintenance will be more when compared to the breakdown maintenance cost. The total cost, which is the sum of the preventive maintenance cost and the breakdown maintenance cost, will go on decreasing with an increase in the level of maintenance up to a point.
Beyond that point, the total cost will start increasing. The level of maintenance corresponding to the minimum total cost is the optimal level of maintenance. The concepts are demonstrated in Fig. A typical shape of each of the above costs with respect to life of the machine is shown in Fig. Replacement and Maintenance Analysis From Fig.
From the beginning, the total cost continues to decrease up to a particular life and then it starts increasing. The point where the total cost is minimum is called the economic life of the machine. If the interest rate is more than zero per cent, then we use interest formulas to determine the economic life. The replacement alternatives can be evaluated based on the present worth criterion and annual equivalent criterion. The basics of these criteria are already presented in Chapter 3.
Based on experience, it was found that the maintenance cost is zero during the first year and it increases by Rs. This is summarized in column B of Table 8. Table 8. The value corresponding to any end of year in this column represents the total maintenance cost of using the equipment till the end of that particular year.
For this problem, the average total cost decreases till the end of year 6 and then it increases. Therefore, the optimal replacement period is six years, i. Discount the maintenance costs to the beginning of year 1. Replacement and Maintenance Analysis 3.
Find Column F by adding the first cost of Rs. Find the annual equivalent total cost through the years given. Identify the end of year for which the annual equivalent total cost is minimum. For this problem, the annual equivalent total cost is minimum at the end of year 7. Therefore, the economic life of the equipment is seven years. End of year Operation cost Maintenance Salvage value n at the end of cost at the at the end of year Rs.
Therefore, the economic life of the machine is five years. The annual equivalent total cost corresponding to the economic life is Rs. Now, the manufacturer of machine B has approached the company. Machine B, which has the same capacity as that of machine A, is priced at Rs.
The maintenance cost of machine B is estimated at Rs. Assume that the scrap value of each of the machines is negligible at any year. Determination of economic life and corresponding annual equivalent total cost of machine B. The details of machine B are summarized in Table 8. The first cost of machine B is equal to Rs. Hence the economic life of machine B is 8 years and the corresponding annual equivalent total cost is Rs. Selection of the best machine is based on the minimum annual equivalent total cost.
The comparison is made over the minimum common multiple of the lives of machine A and machine B, i. In this analysis, the annual equivalent cost of each alternative should be computed first. Then the alternative which has the least cost should be selected as the best alternative.
Before discussing details, some preliminary concepts which are essential for this type of replacement analysis are presented. P Fig. Assume that an equipment has been purchased about three years back for Rs.
The supplier of the new equipment will take the old one for some money, say, Rs. This should be treated as the present value of the existing equipment and it should be considered for all further economic analysis. The purchase value of the existing equipment before three years is now known as sunk cost, and it should not be considered for further analysis.
Its salvage value at the end of its life is Rs. The annual maintenance cost is Rs. The market value of the present machine is Rs. Now, a new machine to cater to the need of the present machine is available at Rs. Its annual maintenance cost is Rs.
The salvage value of the new machine is Rs. It has a present realizable market value of Rs. If kept, it can be expected to last five years more, with operating and maintenance cost of Rs. This engine can be replaced with an improved version costing Rs. This improved version will have an estimated annual operating and maintenance cost of Rs. Equal lives are nothing but the least common multiple of the lives of the alternatives.
Since the annual equivalent cost of the old diesel engine is less than that of the new diesel engine, it is suggested to keep the old diesel engine.
Here, an important assumption is that the old engine will be replaced four times during the 20 years period of comparison. Reinforcement would cost Rs. If it is reinforced, it is estimated that its net salvage value would be Rs. The new prestressed concrete bridge would cost Rs. Such a bridge would have no salvage value.
It is estimated that the annual maintenance cost of the reinforced bridge would exceed that of the concrete bridge by Rs. If the bridge is replaced by a new prestressed concrete bridge, the scrap value of the steel would exceed the demolition cost by Rs. What would you recommend? Solution There are two alternatives: Reinforce the existing bridge. Replace the existing bridge by a new prestressed concrete bridge.
The cash flow diagram for alternative 2 is shown in Fig. Based on equal lives comparison over 40 years, alternative 2 is selected as the best alternative. Thus, it is suggested to go in for prestressed concrete bridge.
Its useful life was estimated to be 10 years. Due to the fast development of that locality, the municipality is unable to meet the current demand for water with the existing motor. The municipality can cope with the situation either by augmenting an additional 5 hp motor or replacing the existing 10 hp motor with a new 15 hp motor.
The details of these motors are now tabulated. Replacement and Maintenance Analysis Solution There are two alternatives to cope with the situation: Augmenting the present 10 hp motor with an additional 5 hp motor. Replacing the present 10 hp motor with a new 15 hp motor. Therefore, it is suggested that the present 10 hp motor be augmented with a new 5 hp motor. Its life is six years and its salvage value at the end of its life is Rs. Now, a company is offering a new machine at a cost of Rs.
Its life is four years and its salvage value at the end of its life is Rs. The annual maintenance cost of the new machine is Rs. The company which is supplying the new machine is willing to take the old machine for Rs.
Solution Old machine Let the comparative use value of the old machine be X. X — 1, 0. Therefore, it is advisable to replace the old machine with the new one.
The failure of the item may result in complete breakdown of the system. The system may contain a collection of such items or just one item, say a tubelight. Therefore, we use some replacement policy for such items which would avoid the possibility of a complete breakdown. The following are the replacement policies which are applicable for this situation.
Under this policy, an item is replaced immediately after its failure. Under this policy, the following decision is made: At what equal intervals are all the items to be replaced simultaneously with a provision to replace the items individually which fail during a fixed group replacement period?
Replacement and Maintenance Analysis There is a trade-off between the individual replacement policy and the group replacement policy. Hence, for a given problem, each of the replacement policies is evaluated and the most economical policy is selected for implementation. This is explained with two numerical problems. If all the transistors are replaced simultaneously, it would cost Rs.
Any one of the following two options can be followed to replace the transistors: Find out the optimal replacement policy, i. If group replacement policy is optimal, then find at what equal intervals should all the transistors be replaced. Solution Assume that there are transistors in use. Let, pi be the probability that a transistor which was new when placed in position for use, fails during the ith week of its life. Assume that a transistors that fail during a week are replaced just before the end of the week, and b the actual percentage of failures during a week for a sub-group of transistors with the same age is same as the expected percentage of failures during the week for that sub-group of transistors.
Replacement and Maintenance Analysis Table 8. From Table 8. Hence, the group replacement period is four weeks. When any resistor fails, it is replaced. The cost of replacing a resistor individually is Rs.
If all the resistors are replaced at the same time, the cost per resistor is Rs. The per cent surviving, S i at the end of month i is tabulated as follows: Solution Let pi be the probability of failure during the month i. Hence, a resistor which has survived for five months would certainly fail during the sixth month.
We assume that the resistors failing during a month are accounted at the end of the month. Replacement and Maintenance Analysis From Table 8. Thus, the group replacement period is three months.
List and explain the different types of maintenance. Discuss the reasons for replacement. Distinguish between breakdown maintenance and preventive maintenance. A firm is considering replacement of an equipment, whose first cost is Rs.
The following table gives the operation cost, maintenance cost and salvage value at the end of every year of a machine whose purchase value is Rs. A manufacturer is offered two machines A and B.
A is priced at Rs. Machine B which has the same capacity is priced at Rs. The maintenance costs of the machine B are estimated at Rs.
Three years back, a machine was purchased at a cost of Rs. Its salvage value at the end of its estimated life is Rs. A new machine to cater to the need of the present machine is available at Rs. A steel highway bridge must either be reinforced or replaced. Three years back, a municipality purchased a 10 hp motor for pumping drinking water.
Its annual operation and maintenance cost is Rs. Due to rapid development of that locality, the municipality is unable to meet the current demand for water with the existing motor. The details of these motors are given in the following table. The failure rates of transistors in a computer are summarized in the following table.
End of week 1 2 3 4 5 6 7 Probability of 0. Find out which is the optimal replacement policy, i. If the group replacement policy is optimal, then find at what equal intervals should all the transistors be replaced. An electronic equipment contains 1, resistors. The per cent surviving, S i at the end of month i is tabulated now. This may be due to wear and tear of the equipment or obsolescence of technology. Hence, it is to be replaced at the proper time for continuance of any business.
The replacement of the equipment at the end of its life involves money. This must be internally generated from the earnings of the equipment. The recovery of money from the earnings of an equipment for its replacement purpose is called depreciation fund since we make an assumption that the value of the equipment decreases with the passage of time.
These are as follows: Straight line method of depreciation 2. Declining balance method of depreciation 3. Sum of the years—digits method of depreciation 4. Sinking-fund method of depreciation 5. Service output method of depreciation These are now discussed in detail. Here, we make an important assumption that inflation is absent. The formulae for depreciation and book value are as follows: The estimated salvage value of the equipment at the end of its lifetime is Rs.
Determine the depreciation charge and book value at the end of various years using the straight line method of depreciation. The calculations pertaining to Bt for different values of t are summarized in Table 9. Table 9. In this approach, it should be noted that the depreciation is the same for all the periods. This approach is a more realistic approach, since the depreciation charge decreases with the life of the asset which matches with the earning potential of the asset.
The book value at the end of the life of the asset may not be exactly equal to the salvage value of the asset. This is a major limitation of this approach. If this rate is used, then the corresponding approach is called the double declining balance method of depreciation. The rates of depreciation for the years 1—8, respectively are as follows: For any year, the depreciation is calculated by multiplying the corresponding rate of depreciation with P — F. The calculations of Dt and Bt for different values of t are summarized in Table 9.
The loss in value of the asset P — F is made available an the form of cumulative depreciation amount at the end of the life of the asset by setting up an equal depreciation amount A at the end of each period during the lifetime of the asset. If we calculate Dt and Bt for all the periods, then the tabular approach would be better. For example, the calculations of net depreciation for some periods are as follows: The minor difference is due to truncation error.
In such cases, the depreciation is computed based on service rendered by an asset. Then, the depreciation is defined per unit of service rendered: Its salvage value after five years is Rs.
The length of road that can be laid by the machine during its lifetime is 75, km. In its third year of operation, the length of road laid is 2, km.
Find the depreciation of the equipment for that year. Define the following: Distinguish between declining balance method of depreciation and double declining balance method of depreciation. The Alpha Drug Company has just purchased a capsulating machine for Rs. Compute the depreciation schedule for the machine by each of the following depreciation methods: A company has recently purchased an overhead travelling crane for Rs.
Its expected life is seven years and the salvage value at the end of the life of the overhead travelling crane is Rs.
Using the straight line method of depreciation, find the depreciation and the book value at the end of third and fourth year after the crane is purchased. An automobile company has purchased a wheel alignment device for Rs. The device can be used for 15 years. Find the following using the double declining balance method of depreciation: A company has purchased a bus for its officers for Rs. The expected life of the bus is eight years.
The salvage value of the bus at the end of its life is Rs.
Find the following using the sinking fund method of depreciation: Consider Problem 4 and find the following using the sum-of-the-years- digits method of depreciation: A company has purchased a Xerox machine for Rs. The salvage value of the machine at the end of its useful life would be insignificant.
The maximum number of copies that can be taken during its lifetime is 1,00,00, During the fourth year of its operation, the number of copies taken is 9,00, Find the depreciation for the fourth year of operation of the Xerox machine using the service output method of depreciation.
A heavy construction firm has been awarded a contract to build a large concrete dam. The firm will buy Rs. During the preparation of the job cost estimate, the following utilization schedule was computed for the special equipment: Prepare the depreciation schedule for all the years of operation of the equipment using the service output method of depreciation.
But the same criterion cannot be used while evaluating public alternatives. Examples of some public alternatives are constructing bridges, roads, dams, establishing public utilities, etc. In this process, one should see whether the benefits of the public activity are at least equal to its costs. If yes, then the public activity can be undertaken for implementation.
Otherwise, it can be cancelled. For the purpose of comparison, these are to be converted into a common time base present worth or future worth or annual equivalent. Similarly, the costs consist of initial investment and yearly operation and maintenance cost. These are to be converted to a common time base as done in the equivalent benefits. If this ratio is at least one, the public activity is justified; otherwise, it is not justified. This results in excessive travel time and increased fuel cost.
So, the state government is planning to construct a bridge across the river. The estimated initial investment for constructing the bridge is Rs. The estimated life of the bridge is 15 years. The annual operation and maintenance cost is Rs. The value of fuel savings due to the construction of the bridge is Rs.
In addition to the production of electric power, this project will provide flood control, irrigation and recreation benefits. The estimated benefits and costs that are expected to be derived from this project are as follows: Project A1 requires an initial outlay of Rs.
The initial outlay for the project A2 is Rs. There is no salvage value associated with either of the projects. Using the benefit cost ratio, which project would you select? Hence, alternative A2 is to be selected. The comparison is made on a year period which is the minimum common multiple of the lives of alternatives 1 and 2. The annual goods transported is 1,00,00, ton km.
The average transport charge is Rs. Within the next 20 years, the transport is likely to increase by 10,00, ton km per year.
It is proposed to broaden a river flowing from the state to the seaport at a cost of Rs. This will make the river navigable to barges and will reduce the transport cost to Rs. There would be some side effects of the change-over as follows. The railroad would be bankrupt and be sold for no salvage value.
The right of way, worth about Rs. The state will have to pay to each of them a welfare cheque of Rs. The reduction in the income from the taxes on the railroad will be compensated by the taxes on the barges. What is the benefit-cost ratio based on the next 20 years of operation? Also, check whether broadening the river is justified.
Since the BC ratio is more than 1, the project is justified. The estimated benefits and costs that are expected from the three alternatives under consideration are given in the following table. Initial cost P 15,00,00, 25,00,00, 40,00,00, 2. Flood control savings 25,00, 35,00, 50,00, 7.
Irrigation benefits 35,00, 45,00, 60,00, 8. Recreation benefits 10,00, 20,00, 35,00, 9. Benefit-cost ratio 1. Since A is the only eligible alternative, it is selected as the best alternative for implementation.
Discuss the difference in evaluating alternatives of private and public organizations. Consider the evaluation of the alternative of constructing a bridge across a river. List the different benefits and costs related to this alternative. In a particular locality of a state, presently, the vehicle users take a roundabout route to reach certain places because of the presence of a river.
This results in excessive time of travel and increased fuel cost. A state government is planning a hydroelectric project for a river basin. The estimated benefits and costs that are expected to be derived from this project are listed below.
Two mutually exclusive projects are being considered for investment. Using the BC ratio, which project would you select? An inland state is presently connected to a seaport by means of a railroad system.
The annual goods transported amount to 1,50,00, ton kilometre. Within the next 25 years, the transport is likely to increase by 15,00, ton kilometre per year. It is proposed to improve a river flowing from the state to the seaport at a cost of Rs.
What is the BC ratio based on the next 25 years of operation? A government is planning a hydroelectric project for a river basin. Besides the production of electric power, this project will provide flood control, irrigation and recreation benefits. The estimated benefits and costs expected from the three alternatives under consideration are listed in the following table: If the rate of inflation is very high, it will produce extremely serious consequences for both individuals and institutions.
Inflation is the rate of increase in the prices of goods per period. So, it has a compounding effect. The same is true for succeeding years and hence the rate of inflation is compounded in the same manner that an interest rate is compounded.
If economic decisions are taken without considering the effect of inflation into account, most of them would become meaningless and as a result the organizations would end up with unpredictable return.
But there is always difficulty in determining the rate of inflation. But due to various reasons, it is very difficult to have zero inflation. For practical decision making, an average estimate may be assumed depending on the period of the proposals under consideration. Hence, we need a procedure which will combine the effects of inflation rate and interest rate to take realistic economic decision. As per our requirement, calculate either the annual equivalent amount or future amount or present amount of the cash flow resulting from step 2 by considering the time value of money.
He plans to retire at the age of 60 and estimates that he can live comfortably on Rs. What equal amount should he save each year until he retires so that he can make withdrawals at the end of each year commencing from the end of the 21st year from now that will allow him to live as comfortably as he desires for 10 years beyond his retirement? Solution Step 1. Step 2. Modification of the costs estimated in step 1 is summarized in Table The formula which is given below is used to get future equivalent of Rs.
Now, the calculation of the equivalent amount of cash flow as per the requirement is presented. The overall cash flow diagram for the savings and withdrawal in terms of future rupees is shown in Fig. Inflation Adjusted Decisions The sum of the present equivalents of the year end withdrawals from the year 21 to 30 is computed by assuming the end of the year 20 as the base time zero and it is shown at the end of the year 20 in Fig.
The method of computing the present equivalent of the withdrawals is as follows: The annual equivalent amount A , which should be invested from the end of year 1 age 41 to year 20 age 60 , is computed using the following formula. The person has to invest an amount of Rs. The productivity of any organization is a function of many factors. It is largely affected by efficient and effective use of machinery and equipment. So, operations and maintenance of these equipment are very important to the organization.
A machine which is purchased today cannot be used forever. It has a definite economic lifetime. After the economic life, the machine should be replaced with a substitute machine with similar operational capabilities. This kind of analysis is called replacement analysis. Purchase cost initial cost 2. Annual operation and maintenance cost 3. Salvage value at the end of every year, if it is significant The trade-off between different cost elements is shown in Fig.
But the capital recovery with return decreases with the life of the machine. The total cost of the machine goes on decreasing initially but it starts increasing after some years.
The year with the minimum total cost is called as the economic life of the machine. Initial cost 2. June Trending on EasyEngineering. December Research, Applications and Advances By K. Gupta Book April An Introduction By Hamdy A.
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